Question

May be I am too tired.
Why dont’t the following display the same value?

int x = 42405;
System.out.println(x << 8);  
System.out.println((x &0x00ff) << 8);  

The lower bits should be clear in both cases

Answer 1

EDIT: Okay, I’m leaving the bottom part for posterity…

If x is int, then it’s pretty simple: x << 8 will have the same value as x & 0xff if and only if none of the “middle” 16 bits are set:

• The top 8 bits of x will be rendered irrelevant by the left shift
• The bottom 8 bits of x are preserved by the masking

If there are any of the “middle” 16 bits set, then x & 0xff will differ from x in a bit which is still preserved by the shift, therefore the results will be different.

I don’t understand why you’d expect them to always give the same results…

---

I’m going to assume that the type of x is byte, and that it’s actually negative.

There’s no shift operation defined for byte, so first there’s a transformation to int… and that’s what can change depending on whether or not you’ve got the mask.

So let’s take the case where x is -1. Then (int) x is also -1 – i.e. the bit pattern is all 1s. Shift that left by 8 bits and you end up with a bit pattern of 24 1s followed by 8 0s:

11111111111111111111111100000000 = -256

Now consider the second line of code – that’s taking x & 0xff, which will only take the bottom 8 bits of the promoted value of x – i.e. 255. You shift that left by 8 bits and you end up with 16 0s, 8 1s, then 8 0s:

00000000000000001111111100000000 = 65280